y^{''}-2y^'+2y=e^{2x}(cos(x)+sin(x))

解

y^{^{''}} - 2 y^{^{'}} + 2 y = e^{2 x} (cos (x) + sin (x))

y = e^{x} (c_{1} cos (x) + c_{2} sin (x)) + \frac{3}{5} e^{2 x} sin (x) - \frac{1}{5} e^{2 x} cos (x)

解答ステップ

y^{''} (x) - 2 y^{'} (x) + 2 y = e^{2 x} (cos (x) + sin (x))

線形 ODE を解く:

y = e^{x} (c_{1} cos (x) + c_{2} sin (x)) + \frac{3}{5} e^{2 x} sin (x) - \frac{1}{5} e^{2 x} cos (x)

y = e^{x} (c_{1} cos (x) + c_{2} sin (x)) + \frac{3}{5} e^{2 x} sin (x) - \frac{1}{5} e^{2 x} cos (x)